China Girls' Mathematical Olympiad

Let $G$ and $M$ be the midpoints of segments $AF$ and $AC$ respectively.

In right triangle $ADM$, $\angle ADM=60^{\circ }$ and $AM=\sqrt{3}DM$.

Note that $AM=3GM$. Hence $DM=\sqrt{3}GM$ and $DG=2GM=AG=GF$. By symmetry, $DF=GF$ and $DFG$ is an isosceles triangle.



In triangle $ADG$, $AG=DG$ and $\angle ADG=30^{\circ }$. Since $\angle ABD=\angle DGF=60^{\circ }$, $ABDG$ is concyclic, implying that $\angle ABG=\angle ADG=30^{\circ }$. Note that $EF$ and $EM$ are midlines in triangles $BGC$ and $BAC$ respectively. In particular, $EF//BG$ and $EM//BA$, implying that $\angle MEF=\angle ABG=30^{\circ }$.

Therefore, $\angle MEF=\angle MDF=30^{\circ }$ and $MDEF$ is concyclic, from which it follows that $\angle DEF=180^{\circ }-\angle DMF=90^{\circ }$.

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Alternative solution.

(We maintain the notations of the first proof.)

Let $N$ be the midpoint of segment $CD$. Then $EN$ and $ME$ are the respective midlines in triangles $BDC$ and $ABC$. In particular, $EN//BD$ and $EM//BA$, implying that $\angle MEN=\angle ABD=60^{\circ }$. Thus, $\angle MDN=\angle MEN=60^{\circ }$, that is, $MDEN$ is concyclic.

It is easy to compute that $AC=\sqrt{3}CD$, $CF=\sqrt{3}CN$ and $\frac{CD}{CM}=\frac{CF}{CN}$. Thus, triangles $CNF$ and $CMD$ are similar. Consequently, we deduce that $\angle CNF=\angle FMD=90^{\circ }$, that is, $MDNF$ is concyclic.



Therefore, $MDENF$ is concyclic (with $DF$ as its diameter) and $\angle DEF=90^{\circ }$.